Answer:
0.321 M
Explanation:
The neutralization reaction of HCl with NaOH is the following:
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
So, 1 equivalent of HCl reacts with 1 equivalent of NaOH. In this case:
1 equivalent HCl = 1 mol HCl
1 equivalent NaOH = 1 mol NaOH
At the equivalence point, the amount of HCl reacts completely with the added amount of titrant (NaOH). As 1 mol of reactant is calculated from the product of the molarity of the solution and the volume, in the equivalent point:
moles HCl = moles NaOH
M(HCl) x V(HCl) = M(NaOH) x V(NaOH)
We have:
M(NaOH) = 0.150 M
V(NaOH) = 21.4 mL = 0.0214 L
V(HCl) = 10.0 mL = 0.01 L
Thus, we calculate the initial concentration HCl as follows:
M(HCl) = (M(NaOH) x V(NaOH))/ (V(HCl))
= (0.150 M x 0.0214 L)/(0.01 L) = 0.321 M